A 1.00M solution of a hypothetical weak acid (HB) has a pH of 2.85. A solution is created using 0.10M of a salt NaB ('B' being the anion of the weak acid HB) and 50.0 ml of 2.0M acid HB (assume no change in volume). What would be the pH of the solution and what would you call the solution?
Use the 1 M HB and pH of 2.85 to determine pKa for HB. The remaining part of the BUFFER problem is solved by using the Henderson-Hasselbalch equation.
To calculate the pH of the solution, we need to consider the dissociation of the weak acid HB and the hydrolysis of the salt NaB.
Step 1: Calculate the initial concentration of HB in the solution.
Since we have a 50.0 mL solution of 2.0M acid HB, the moles of HB can be calculated as follows:
moles of HB = concentration × volume = 2.0 M × 0.0500 L = 0.100 mol
Since the volume doesn't change, the final concentration of HB in the solution is also 0.100 mol/L.
Step 2: Calculate the concentration of NaB.
The concentration of NaB is given as 0.10M.
Step 3: Evaluate the hydrolysis reaction of NaB.
NaB dissociates in water to form Na+ ions and the anion B-. The B- ions can react with water to produce hydroxide ions (OH-) and regenerate the weak acid HB.
The hydrolysis reaction is as follows:
B- + H2O ⇌ HB + OH-
Since NaB is a salt of the weak acid HB, we assume complete ionization and use the stoichiometry to calculate the concentration of OH- ions produced.
0.10 M of NaB would produce 0.10 M of B- ions.
Therefore, the concentration of OH- ions would also be 0.10 M.
Step 4: Calculate the concentration of HB after hydrolysis.
Since HB is a weak acid, it does not completely dissociate. Therefore, the equilibrium concentration of HB can be calculated using the initial concentration and the concentration of OH- ions:
[H+][OH-] / [HB] = (10^(-pH))(0.10) / [HB] = Kw = 1.0 x 10^(-14)
To solve for [HB], we need to know the value of the equilibrium constant (Kw), which represents the product of the concentrations of H+ and OH- ions in water, assuming neutral conditions. Kw is approximately 1.0 x 10^(-14) at 25°C.
Step 5: Calculate the pH of the solution.
Using the given pH of the 1.00M solution as a reference, we can use the relationship between concentration and pH to find the pH of the new solution.
pH = -log10[H+]
[H+] = 10^(-pH)
Since [H+] is inversely proportional to [HB], we can calculate [H+] as follows:
[H+] = (10^(-pH))(0.10 mol/L) / [HB]
Substituting the value of [HB] from Step 4:
[H+] = (10^(-pH))(0.10 mol/L) / (0.100 M - 0.10 mol/L)
Now, you can substitute the given pH value (pH = 2.85) into the equation to calculate the concentration of H+ ions.
Finally, once you have the concentration of H+ ions, you can convert it back to pH using the equation pH = -log10[H+].
As for what you would call the solution, it depends on the pH value obtained. pH values below 7 are acidic, pH values equal to 7 are neutral, and pH values above 7 are basic.